## Miscellaneous Science Questions

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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

My question has probably been answered and will most likely only show my ineptitude in the field, but I was thinking thought you all might be able to clear something up for me.
Heisenburg's uncertainty principle is of course m(delta)v(delta)x>h
If a photon is travelling in a vacuum at c, and if it in fact has no resting mass, then the equation would not be greater than Planck's Constant. Also, the fact that it was in fact travelling at c would cause the uncertainty of the velocity to be zero, also cancelling the equation. This implies that either a light particle can violate Heisenburg's uncertainty principle when operating inside of a vacuum, or the there is an equivalent to zero point motion for both mass and velocity. I also could be missing something entirely. Someone please come to my aid!
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### Re: RELATIVITY QUESTIONS! (and other common queries)

In relativity, momentum is found using
E^2 = p^2 c^2 + m^2 c^4
m=0 for a photon, but E does not! Its momentum doesn't depend on its velocity, just its energy (which is to say, wavelength/frequency). So then the uncertainty for a photon becomes even less strange looking, now it is a statement about not knowing position and wavelength simultaneously. It's just like regular spectral broadening. Since p = h/[imath]\lambda[/imath], it now says
$\Delta\lambda\Delta x > 1$
And there might be some 2pi's, but whatever.
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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

But how does the photon end up with mass? even with energy and momentum, wouldn't 0mass still cancel the equation?

(Also, a bit off topic, but how do you type delta on a PC?)

And thank you
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### Re: RELATIVITY QUESTIONS! (and other common queries)

wbbaxterbones wrote:My question has probably been answered and will most likely only show my ineptitude in the field, but I was thinking thought you all might be able to clear something up for me.
Heisenburg's uncertainty principle is of course m(delta)v(delta)x>h

This is your problem. It's not m(delta)v. It's (delta)p, where p is momentum. You can't break that down into m(delta)v without making a few assumptions and redefinitions to make it fit with relativity, as you found.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

You go into math or imath mode, by typing inside {math}{\math}, but with [ instead of {. Or click the button on top to put you in the mode.
Then, once you're in the mode, you do \delta for lowercase and \Delta for upper. It's TeX style.
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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Wait, don't you need inertia to have momentum, and inertia is relative to mass? Therefore wouldn't an object need mass to generate momentum?
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### Re: RELATIVITY QUESTIONS! (and other common queries)

No, because momentum in relativity is not simply mass times velocity.

For massive and massless things, momentum is [imath]h/\lambda[/imath], where [imath]\lambda[/imath] is the wavelength.
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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

One more question, hopefully.

If E^2=p^2c^2+m0^2c^4 such that dE/dp=pc^2/E=v

following that E=mc^2=m0c^2/(1-v^2/c^2)^(1/2); and p=mv=m0v/(1-v^2/c^2)^(1/2)

Wouldn't this imply that in the case of a photon with 0 resting mass(0m0), the velocity will be undefined or 0, while the momentum would also be 0, thus rendering Heisenberg's Uncertainty Principle worthless to define its behaviours?

This is the case in which I am confused, but there is probably some glaring concept I have missed that you could point out. I apologize if you had trouble reading my equation. I can't figure out math TeX.
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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

o, wait!

do you use E=ħω and p=ħk with non-relativistic particles?
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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

I am sorry, but I don't see how that answers my question insofar as how a photon can follow heisenberg's uncertainty principle. Thanks anyway though!
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Talking about the rest mass of a photon or other "massless" particle can get tricky. Use the expressions for momentum and energy, and leave [imath]E=\gamma m_0c^2[/imath] out of it.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

[imath]E^2 = p^2c^2+m^2c^4[/imath] is fine. It is [imath]p=\gamma m v[/imath] that is a problem.

wbbaxterbones wrote:o, wait!
do you use E=ħω and p=ħk with non-relativistic particles?

You certainly do.

Special relativity and quantum mechanics don't actually play together nicely on their own, you have to go to quantum field theory if you want things to make sense in a precise way.
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wbbaxterbones
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Thanks to everyone who answered. You were very helpful, and I am hopefully starting developing an understanding of the subject matter.
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Xen[Tec]™
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### Re: RELATIVITY QUESTIONS! (and other common queries)

I'll admit that I have limited knowledge of Relativity, Thermodynamics, and (No knowledge) of higher math. However, since (As I understand it) the rate of time is essentially the speed of light minus the object in question's own velocity, wouldn't that mean that time would be in violation of the second law of thermodynamics (Diminishing returns) were it to be circular and/or infinite? Haven't really thought about it much, just been toying with the notion.
Another question, does light have a limit to how long it can exist?

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Xen[Tec]™ wrote:wouldn't that mean that time would be in violation of the second law of thermodynamics (Diminishing returns) were it to be circular and/or infinite?

Interestingly enough, that law is pretty much the only physical theory that differentiates between past and present. Thermodynamics defines an arrow of time. Infinity wouldn't put time in violation of the second law (you just get more and more entropy), but circular arguably would.
Another question, does light have a limit to how long it can exist?

I'm not sure what you mean. A photon, once emitted, continues forever until it hits something.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Sir_Elderberry wrote:
Xen[Tec]™ wrote:wouldn't that mean that time would be in violation of the second law of thermodynamics (Diminishing returns) were it to be circular and/or infinite?

Interestingly enough, that law is pretty much the only physical theory that differentiates between past and present. Thermodynamics defines an arrow of time. Infinity wouldn't put time in violation of the second law (you just get more and more entropy), but circular arguably would.
Another question, does light have a limit to how long it can exist?

I'm not sure what you mean. A photon, once emitted, continues forever until it hits something.

Ah, just what I wanted to know. Thanks

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Sir_Elderberry wrote:
Xen[Tec]™ wrote:wouldn't that mean that time would be in violation of the second law of thermodynamics (Diminishing returns) were it to be circular and/or infinite?

Interestingly enough, that law is pretty much the only physical theory that differentiates between past and present. Thermodynamics defines an arrow of time. Infinity wouldn't put time in violation of the second law (you just get more and more entropy), but circular arguably would.

There's also causality, which isn't usually referred to as a law, but is regarded as a necessary consequence of Special Relativity. Any event that influences some other event must lie in the past light cone of the influenced event.
Another question, does light have a limit to how long it can exist?

I'm not sure what you mean. A photon, once emitted, continues forever until it hits something.

Light is redshifted by the expansion of the Universe. I don't think there is a natural limit to this, and eventually it will fade so much that it becomes undetectable.

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MrGee
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### Consequences of Velocity -> Mass

This is probably a stupid question, but:

Supposedly, if you try to accelerate past the speed of light, your spaceship gets heavier and heavier and you never actually get there. So my question is, if you have a powerful-enough engine, couldn't you eventually put in so much energy that you turn into a black hole (because your mass increases)? Or failing that, could you create some sort of weird gravity fields--like say you do a flyby of Earth at .999999c, would you produce a big gravity field that would tend to lift things off the surface?

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### Re: Consequences of Velocity -> Mass

MrGee wrote:So my question is, if you have a powerful-enough engine, couldn't you eventually put in so much energy that you turn into a black hole (because your mass increases)?

Yes. Increasing your energy density indefinitely will, eventually, cause you to form an event horizon.

That's assuming some sort of magic drive, though. Generally engines work by taking energy you already have and freeing it, which lowers your energy density obviously. Something like a light sail would do it, though, as you're receiving energy from outside.
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### Re: Consequences of Velocity -> Mass

Xanthir wrote:
MrGee wrote:So my question is, if you have a powerful-enough engine, couldn't you eventually put in so much energy that you turn into a black hole (because your mass increases)?

Yes. Increasing your energy density indefinitely will, eventually, cause you to form an event horizon.

That's assuming some sort of magic drive, though. Generally engines work by taking energy you already have and freeing it, which lowers your energy density obviously. Something like a light sail would do it, though, as you're receiving energy from outside.

I wonder, though, if you are using a light sail to absorb photons and thus gain energy, would your energy really increase indefinitely? The thing is, as you accelerate away from the light source, the light will redshift more and more, each photon providing less energy than the last. At the same time, you are becoming more and more massive, thus increasing your inertia. In principle, if you could do this forever, I guess your speed would still approach c, so your mass should approach infinity, and eventually you would become a black hole. But you can't go for infinity, because eventually the wavelengths will become too long for any reasonably-sized sail to absorb. So it doesn't seem that this would be possible for anything driven by a sail.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Your "mass" doesn't have to hit infinity, though. Just "really big".
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### Re: Consequences of Velocity -> Mass

Eebster the Great wrote:
Xanthir wrote:your mass should approach infinity, and eventually you would become a black hole.

Schwarzchild radius and therefore black hole formation depends on rest mass (I made the same mistake earlier in the thread).
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Really? Interesting.
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### Re: Consequences of Velocity -> Mass

eSOANEM wrote:
Eebster the Great wrote:
Xanthir wrote:your mass should approach infinity, and eventually you would become a black hole.

Schwarzchild radius and therefore black hole formation depends on rest mass (I made the same mistake earlier in the thread).

That is somewhat strange.

It implies you can have an infinite energy density using photons without forming a black hole.

... and that seems wrong.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

OK, I have an unrelated question. It's the Heisenberg Uncertainty Principle. Some sources say it is [imath]m \Delta v \Delta x \ge {\hbar \over 2}[/imath], while others say it is [imath]\Delta p \Delta x \ge {\hbar \over 2}[/imath]. But these aren't equivalent--isn't [imath]p = \lambda m v[/imath]? Which is it?

While posting this I just realized that [imath]p \ge m v[/imath], so I guess if the former is true, the latter is also true. So I guess the question is, is [imath]m \Delta v \Delta x \ge {\hbar \over 2}[/imath] necessarily?

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### Re: Consequences of Velocity -> Mass

Yakk wrote:
eSOANEM wrote:
Eebster the Great wrote:
Xanthir wrote:your mass should approach infinity, and eventually you would become a black hole.

Schwarzchild radius and therefore black hole formation depends on rest mass (I made the same mistake earlier in the thread).

That is somewhat strange.

It implies you can have an infinite energy density using photons without forming a black hole.

... and that seems wrong.

Yeah, I remember someone saying something like "there are no 'real' event horizons, only apparent event horizons."
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Eebster, the conversation that I started at the top of the page might answer your question.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

wbbaxterbones wrote:Eebster, the conversation that I started at the top of the page might answer your question.

But Heisenberg still totally matters in QFT, and momentum is still greater than m v, so that doesn't answer my question.

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### Re: Consequences of Velocity -> Mass

eSOANEM wrote:
Eebster the Great wrote:
Xanthir wrote:your mass should approach infinity, and eventually you would become a black hole.

Schwarzchild radius and therefore black hole formation depends on rest mass (I made the same mistake earlier in the thread).

What about the gravity thing? Would observers in any reference frame detect a change in your gravity because you were going so fast?

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Schwarzschild radius also depends on the mass being spherically symmetric and alone in the universe, so it's just an approximation.

Anyway, GR in general depends on the stress-energy tensor iirc, which is not just mass, and not just energy.
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### Re: Consequences of Velocity -> Mass

Yakk wrote:[snip]

As far as I understand it, as long as the photons are all travelling parallel to each other then they will 'outrun' the gravitational attraction they cause. If they are not parallel then I'm not sure.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Hello all.

I still have problems understanding relativity, though I guess that is quite normal.

To start, I have some yes/no and either A or B questions.

As far as I know, there is no such thing in reality as an inertial frame of reference? (Among a multitude of other things in reality that limit our accuracy...sigh)
Anyway, does time dilate and length contract simultaneously, or can you only say that an object has its length contracted and time is 'normal'?
For a mass with infinite energy, it will get from point A in space to point B in space in zero time, right?

Now for non A or B questions
Also, I do not learn anything about Poincare Groups or advance mathematics at my level yet. What does the advance mathematics describe? How much am I crippled in learning relativity? (Starting to learn matrices this year too if I recall my syllabus correctly, -.-)

So anyway, some time ago I asked what velocity a spaceship would need to have to reach alpha centauri(Let me ignore all gravitational effects for now even though obviously there are gravitational effects) in 4.37 years(With the clock of the spaceship), given that it is 4.37 light years away(This value is measured from earth), with uniform motion.

I tried this equation:
vγ=dγ/t

After some simple algebra...
Spoiler:
vγ=(4.37year)c/4.37year |(Yes I wanted my c to come out)
vγ=c
v=c/γ
v=c(sqrt(1-v2/c2))
v2=c2(1-v2/c2)
2v2=c2
v2=c2/2
v=c/sqrt 2

v=0.707c

Now comes some more yes/no questions.

Is the distance between earth and alpha centauri given by d/γ(3.09 ly according to google), if the spaceship were to measure it?
Is the velocity v relative to the Earth?(Yes I am asking) Can one consider the earth as an inertial frame of reference?
Does the spaceship reach alpha centauri after tγ(6.17 years according to google) has passed, according to Earth?

Somehow, I am getting all of these numbers simply by 'logic' and 'deduction,' rather than true knowledge of what relativity is about, using 'laws' like 'nothing with mass goes as fast as light.'

Finally, I want to calculate v better. Given dγ and t, or distance relative to earth and time that would pass on the ship, what would v be for any distance dγ?
If I wanted to, I probably could plug in 10 billion light years and 3.14 seconds and get an answer using
vγ=dγ/t, as above.
However, I am unable to express v in terms of dγ, t, and c, which are all known. The resulting equation I get is
v=sqrt ((dγ)2/t2+c2)
Which is very wrong.

My working is:
Spoiler:
vγ=dγ/t
v=(dγ/t)sqrt(1-v2/c2)
v2=((dγ)2/t2)(1-v2/c2)
v2=(dγ)2/t2 - v2(dγ)2/c2t2
v2+v2(dγ)2/c2t2=(dγ)2/t2
v2(1+(dγ)2/c2t2)=(dγ)2/t2
v2=((dγ)2/t2)/(1+(dγ)2/c2t2
v2=(dγ)2/t2+c2
v=sqrt ((dγ)2/t2+c2)

So what did I do wrong?

And no, all of this is not homework, seeing as vγ doesn't seem to make a lot of physical sense. (Though it pains me that I am taught that EM waves are 'waves and not particles.' Not teaching wave-corpuscle duality is not very cool)

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### Re: RELATIVITY QUESTIONS! (and other common queries)

QwertyKey wrote:As far as I know, there is no such thing in reality as an inertial frame of reference? (Among a multitude of other things in reality that limit our accuracy...sigh)

I'm not sure what you mean by this. There is no single absolute reference frame, but any reference frame can be considered inertial.

Anyway, does time dilate and length contract simultaneously, or can you only say that an object has its length contracted and time is 'normal'?

In special relativity (SR), time dilation and length contraction are phenomena experienced by observing motion relative to the observer's reference frame, and both always occur. That is, if I am in the Earth's reference frame, and I watch a train go by at 0.99 c, I will see the clock on the train running very slowly, and the train will look very short. However, an observer on the train will see my clock moving very slowly, and I will look very skinny (unless I am lying down horizontally, and then I will look very short).

In general relativity (GR), spacetime can also be warped by the presence of mass, and this is the cause of gravity. In a sense, these effects are the result of relative acceleration of reference frames, rather than relative motion.

For a mass with infinite energy, it will get from point A in space to point B in space in zero time, right?

That sentence doesn't really make sense. First of all, for a body to have infinite energy, it must either have infinite mass (which is obviously impossible, as it would form a black hole long before) or have nonzero mass and be moving at the speed of light, which is also impossible, but even if this were the case, it would be moving at the speed of light, and no faster. For a body with nonzero, finite mass (and therefore nonzero, finite energy), the speed of light is an absolute speed limit. A body with zero mass (e.g. a photon) will always move at exactly the speed of light. (We would certainly have a problem if light didn't move at the speed of light!)

Also, I do not learn anything about Poincare Groups or advance mathematics at my level yet. What does the advance mathematics describe? How much am I crippled in learning relativity? (Starting to learn matrices this year too if I recall my syllabus correctly, -.-)

You can understand SR to some extent with only basic calculus and algebra and some understanding of vectors. That's what I did, at least. You will need to understand matrices, though. GR is of course more complicated, and I assume it requires analytic geometry at the very least.

But in either case, you can understand SR and GR conceptually without any precise mathematical understanding. In fact, Einstein did not solve most of his own equations (the nontrivial differential equations). SR and GR make a lot of sense just from thinking about Einstein's "thought experiments" without much (or any) math. But you can't really apply this conceptual knowledge in any practical way without the math.

So anyway, some time ago I asked what velocity a spaceship would need to have to reach alpha centauri(Let me ignore all gravitational effects for now even though obviously there are gravitational effects) in 4.37 years(With the clock of the spaceship), given that it is 4.37 light years away(This value is measured from earth), with uniform motion.

I tried this equation:
vγ=dγ/t

After some simple algebra...
Spoiler:
vγ=(4.37year)c/4.37year |(Yes I wanted my c to come out)
vγ=c
v=c/γ
v=c(sqrt(1-v2/c2))
v2=c2(1-v2/c2)
2v2=c2
v2=c2/2
v=c/sqrt 2

v=0.707c

I don't understand all your steps exactly, but your answer looks correct. However, let's use x for distance instead of d, because d will get confused for a differential.

Is the distance between earth and alpha centauri given by d/γ(3.09 ly according to google), if the spaceship were to measure it?

Yes, the spaceship you described will see Alpha Centauri as being 3.09 ly away at the beginning of its trip, while the Earth will see it as being 4.37 ly away.

Is the velocity v relative to the Earth?(Yes I am asking) Can one consider the earth as an inertial frame of reference?

Uh, I was assuming you were defining the speed as being relative to the Earth (or to Alpha Centauri, since they are approximately at rest). I don't know how else you would define it that makes sense. The Earth is not an inertial reference frame, technically, since it is accelerating (rotating and orbiting the sun). However, these are very slight relative to the speed of the spaceship, so you can consider the Earth to be approximately in an inertial reference frame for the same reason you can consider it to be approximately at rest with respect to Alpha Centauri. (Technically, all reference frames can be considered inertial when varying gravitational fields are added, but this is not simple or necessary in this case.)

Does the spaceship reach alpha centauri after tγ(6.17 years according to google) has passed, according to Earth?

Yes, although an even simpler calculation is t = x/v = 4.37 ly / (c / sqrt 2) = 6.18 ly (you must have made a rounding error).

Somehow, I am getting all of these numbers simply by 'logic' and 'deduction,' rather than true knowledge of what relativity is about, using 'laws' like 'nothing with mass goes as fast as light.'

You should think of each step entirely within a given reference frame if possible, and then apply the Lorentz Transformations.

Finally, I want to calculate v better. Given dγ and t, or distance relative to earth and time that would pass on the ship, what would v be for any distance dγ?
If I wanted to, I probably could plug in 10 billion light years and 3.14 seconds and get an answer using
vγ=dγ/t, as above.
However, I am unable to express v in terms of dγ, t, and c, which are all known. The resulting equation I get is
v=sqrt ((dγ)2/t2+c2)
Which is very wrong.

My working is:
Spoiler:
vγ=dγ/t
v=(dγ/t)sqrt(1-v2/c2)
v2=((dγ)2/t2)(1-v2/c2)
v2=(dγ)2/t2 - v2(dγ)2/c2t2
v2+v2(dγ)2/c2t2=(dγ)2/t2
v2(1+(dγ)2/c2t2)=(dγ)2/t2
v2=((dγ)2/t2)/(1+(dγ)2/c2t2
v2=(dγ)2/t2+c2
v=sqrt ((dγ)2/t2+c2)

So what did I do wrong?

Uh, you really need to define your variables. To me, it looks like you are just using v = d/t, so γ v = γ d/t, where v is speed, d is distance, t is time, and γ is any constant, which is true, but not useful. You should just use the standard <ct, x, y, z> and <ct', x', y', z'> if you don't want to use xμ notation.

Furthermore, I don't know what you mean by v. Do you mean the speed, v = dx/dt, or the proper speed, w = dx/dτ? These are two distinct concepts.

O.K., I just read through again, and it sounds like you want a general solution to the problem you stated above for any distance x' (from the Earth's reference frame). This is simple--just apply the same steps you did before, but don't substitute x = 4.37 ly. You should get v = x/t = γ x'/t, so v/γ = v^2 - v^4/c^2 = x'/t, so
$v = \sqrt{\frac{c (\sqrt{c^2 t - 4x'} + c \sqrt{t})}{2 \sqrt{t}}}$
where x' is the distance from the beginning to the destination from the Earth's reference frame, t is time from the spaceship's reference frame, v is the speed of the spaceship relative to the Earth, and c is the speed of light.

And no, all of this is not homework, seeing as vγ doesn't seem to make a lot of physical sense. (Though it pains me that I am taught that EM waves are 'waves and not particles.' Not teaching wave-corpuscle duality is not very cool)

vγ only makes sense when comparing one reference frame to another, like all speeds. w = dx/dt' = v γ is called the "proper speed."

QwertyKey
Posts: 49
Joined: Tue Aug 04, 2009 4:45 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

Thank you so much for clarifying so many of my doubts.

As far as I know, there is no such thing in reality as an inertial frame of reference? (Among a multitude of other things in reality that limit our accuracy...sigh)

I'm not sure what you mean by this. There is no single absolute reference frame, but any reference frame can be considered inertial.

I thought about my question and I guess what I really wanted to ask is: as far as I know, there is no such thing in reality as a frame of reference of a rest mass, because no mass is ever at rest. Momentum, from what I understand from quantum mechanics, is never equal to zero.

For a mass with infinite energy, it will get from point A in space to point B in space in zero time, right?

That sentence doesn't really make sense. First of all, for a body to have infinite energy, it must either have infinite mass (which is obviously impossible, as it would form a black hole long before) or have nonzero mass and be moving at the speed of light, which is also impossible, but even if this were the case, it would be moving at the speed of light, and no faster. For a body with nonzero, finite mass (and therefore nonzero, finite energy), the speed of light is an absolute speed limit. A body with zero mass (e.g. a photon) will always move at exactly the speed of light. (We would certainly have a problem if light didn't move at the speed of light!)

I get this idea from http://en.wikipedia.org/wiki/Four-velocity. For a non zero mass, if it moves at the speed of light, its clocks will simply stop ticking. By travelling only space, it should reach point B in space from point A in zero time, seeing as time does not move for it.

In other words, the norm or magnitude of the four-velocity is always exactly equal to the speed of light. Thus all objects can be thought of as moving through spacetime at the speed of light.
~ Wiki Page

So all things travel at the speed of light, just that we travel in time, while photons and possible gravitons travel only in the spatial dimensions. As for universal expansion speed, I have no freaking idea, but so far I heard it is faster than c.

Also, would it be possible for a blackhole to move? Just a question I had since you mentioned them.

vγ=dγ/t
is
v for velocity relative to Earth (dx/dt) (I would prefer an observer at rest)
γ is the lorentz factor
d is the distance
t is the time that passes on the spaceship

I simply wanted to calculate v, given only dγ(Your x′) and t(You use t too)

Well, I guess I should use the standard, but I have little idea of matrices.

I also did not realise that vγ is proper speed until now, because my differentiation and integration is very rusty.

Thanks for your formula, I should try it out one day, when I have the time.

Eebster the Great
Posts: 3423
Joined: Mon Nov 10, 2008 12:58 am UTC
Location: Cleveland, Ohio

### Re: RELATIVITY QUESTIONS! (and other common queries)

QwertyKey wrote:Thank you so much for clarifying so many of my doubts.

As far as I know, there is no such thing in reality as an inertial frame of reference? (Among a multitude of other things in reality that limit our accuracy...sigh)

I'm not sure what you mean by this. There is no single absolute reference frame, but any reference frame can be considered inertial.

I thought about my question and I guess what I really wanted to ask is: as far as I know, there is no such thing in reality as a frame of reference of a rest mass, because no mass is ever at rest. Momentum, from what I understand from quantum mechanics, is never equal to zero.

Interesting. I think you can define an inertial reference frame, but you can never measure any object to be at rest relative to it. I guess then it doesn't make sense to define your reference frame according to a single particle, because that particle's momentum is a probability distribution. However, you can define your reference frame based on large objects that approximately define a single reference frame, like the Earth.

For a mass with infinite energy, it will get from point A in space to point B in space in zero time, right?

That sentence doesn't really make sense. First of all, for a body to have infinite energy, it must either have infinite mass (which is obviously impossible, as it would form a black hole long before) or have nonzero mass and be moving at the speed of light, which is also impossible, but even if this were the case, it would be moving at the speed of light, and no faster. For a body with nonzero, finite mass (and therefore nonzero, finite energy), the speed of light is an absolute speed limit. A body with zero mass (e.g. a photon) will always move at exactly the speed of light. (We would certainly have a problem if light didn't move at the speed of light!)

I get this idea from http://en.wikipedia.org/wiki/Four-velocity. For a non zero mass, if it moves at the speed of light, its clocks will simply stop ticking. By travelling only space, it should reach point B in space from point A in zero time, seeing as time does not move for it.

See, here again we have a confusion of variables. From the perspective of the object moving at the speed of light, no time has passed during the trip because points A and B were actually the same point (i.e. the ship did not move). From the perspective of somebody else looking at the object, it moved from point A to point B at the speed of light.

In other words, the norm or magnitude of the four-velocity is always exactly equal to the speed of light. Thus all objects can be thought of as moving through spacetime at the speed of light.
~ Wiki Page

So all things travel at the speed of light, just that we travel in time, while photons and possible gravitons travel only in the spatial dimensions. As for universal expansion speed, I have no freaking idea, but so far I heard it is faster than c.

Yes, the magnitude of the four-velocity is always c. Space expanding is an entirely different matter, but if two objects are far enough away, their distance may be increasing at a rate greater than c.

Also, would it be possible for a blackhole to move? Just a question I had since you mentioned them.

Of course, black holes are just big, dense black bodies. Actually, they are really complicated, but from the perspective of somebody looking at them from a distance, they are just big, dark stars, sometimes spinning, and sometimes with an electric charge.

vγ=dγ/t
is
v for velocity relative to Earth (dx/dt) (I would prefer an observer at rest)
γ is the lorentz factor
d is the distance
t is the time that passes on the spaceship

I simply wanted to calculate v, given only dγ(Your x′) and t(You use t too)

Well, I guess I should use the standard, but I have little idea of matrices.

I also did not realise that vγ is proper speed until now, because my differentiation and integration is very rusty.

Thanks for your formula, I should try it out one day, when I have the time.

I guess that makes sense. In the future, I think it will be less confusing if you use <ct,x,y,z> for the coordinates in one reference frame and <ct',x','y,z'> for the coordinates in another reference frame, or something like that. You don't need to know anything about matrices to do that, and that isn't specific to relativity.

Homer
Posts: 7
Joined: Mon Feb 15, 2010 11:01 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

I've got a question that's been nagging me for years and thought this would be a good place to discuss it.

Suppose you're near a massive black hole. Nothing escapes a black hole but gravity, yes? (- let's ignore virtual particles for simplicity's sake.) Now, let's say you have a transmitter that uses gravity waves instead of radio waves. You grab your transmitter, jump into the black hole and begin broadcasting the sights and sounds. Granted you'll be killed when the tidal forces get big enough but, if the black hole is big enough, that wont happen until you're well past the event horizon.

Meanwhile, in orbit around the black hole, is your partner with a gravity wave detector. Shouldn't he be able to pick up your broadcast, even after you cross the event horizon, and thus effectively "see" inside a black hole?

eSOANEM
:D
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Gravity doesn't escape from within the event horizon. The gravitational field of a black hole, like its electro-magnetic one if it has one, is a skeleton field left behind by the star before it collapsed, due to gravitational time dilation, the field doesn't disappear but remains.

Because gravity can't escape, a gravitational communication device would be just as useless as a radio transmitter.
my pronouns are they

Magnanimous wrote:(fuck the macrons)

Homer
Posts: 7
Joined: Mon Feb 15, 2010 11:01 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

eSOANEM wrote:Gravity doesn't escape from within the event horizon.

Unless I'm very much mistaken, gravity does escape the event horizon i.e. it doesn't begin at that boundary but is continuous through it. Thus the region between the point mass at the center of the black hole and the event horizon is "real".

But, if what you say is right, I can see how the whole thing falls apart.

doogly
Dr. The Juggernaut of Touching Himself
Posts: 5530
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Location: Lexington, MA
Contact:

### Re: RELATIVITY QUESTIONS! (and other common queries)

A static black hole has a gravitational field that extends continuously past the horizon. If you are a source of gravitational disturbances inside the horizon, you cannot send signals across it. It is exactly analagous to e&m waves.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.

Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?

Homer
Posts: 7
Joined: Mon Feb 15, 2010 11:01 pm UTC

### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:If you are a source of gravitational disturbances inside the horizon, you cannot send signals across it. It is exactly analagous to e&m waves.

But a black hole does in effect send out a signal. Consider the gravity as just a carrier wave. I'm suggesting it can be modulated. Since it's not EM in nature it's not going to be pulled back in.