Speed of light and this Accelerating Rocket scenario

In summary, the conversation discussed the link between gravity and the speed of light, focusing on the limitations presented by the Special Theory of Relativity. The participants debated the possibility of exceeding the speed of light through the use of external gravity, but ultimately concluded that this was not possible due to the mathematical framework of SR. The conversation also touched on the concept of relativistic mass and the four-momentum of an object with mass.
  • #1
John SpaceY
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TL;DR Summary
Link between speed and external gravity to a moving mass m (a spacecraft for example)
Hello,
I have a question linked to gravity and speed of light :

According to the Special Theory of Relativity, the speed of light cannot be exceeded because it would need an infinitive energy to accelerate the relativistic mass of a moving objet (a space craft for example) : indeed if the moving object has a mass m at rest, this mass would become M (the relativistic mass) and this mass M will tend towards infinity when the speed v of m will go towards c.
I understand this point, and so if the spacecraft of mass m has a motor, this motor will not be able to give an infinite enrgy to m and so v will be always lower than c.

But I see an exception if we consider an external gravity to m.
Suppose that the motor of the spacecraft could bring enough energy to reach a speed of 0,9c for the speed of m (not possible today but one day maybe) : if there is an external gravity to the mass m, this external gravity (a "black-hole for example with a very high mass MG) will accelerate the mass m towards him and this acceleration is not linked to the mass m of the spacecraft (and so it is not linked to its relativistic mass M).

The acceleration of the external gravity MG will be proportional to the mass MG and will be divided by the squared distance between m and the center of the gravity MG.

And so if the spacecraft of mass m (or M) will be very near of MG and if MG is very high, with only few steps of calculation the speed of the spacecraft will go from 0,9c to c (because of the high acceleration given by the external gravity MG to m and this acceleration is not linked to m and so not limited by the relativistic mass M and so no issue with a lack of energy), and the speed will exceed c very soon.

The situation is different here because the energy doesn't come from inside the spacecraft (from the motor for example) but the energy comes from the outside and the acceleration is not linked to m.

I would like to know if this calculation is OK or if there is a mistake ?
And if there is a mistake, where it is and what is the mistake ?

Thank you in advance for your answers
John
 
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  • #2
It's nonsense. You cannot exceed the speed of light in a relativistic universe, full stop. A thing with non-zero mass travelling at the speed of light is a direct self-contradiction.
 
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  • #3
John SpaceY said:
TL;DR Summary: Link between speed and external gravity to a moving mass m (a spacecraft for example)

I would like to know if this calculation is OK or if there is a mistake ?
And if there is a mistake, where it is and what is the mistake ?
The calculation is wrong because it is using a Newtonian gravity formula in a relativistic context. The correct equation for the relativistic context is the Einstein field equations: https://en.wikipedia.org/wiki/Einstein_field_equations

As @Ibix correctly pointed out, this prevents any thing with mass from traveling at or beyond the speed of light locally. It is firmly set in the mathematical framework. This can be more easily seen from the four-momentum which is easier than dealing with the full EFE.
https://en.wikipedia.org/wiki/Four-momentum
The four-momentum of an object with mass is always timelike, so it cannot travel at c (lightlike) or faster (spacelike).
 
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  • #4
Thanks Dale for the answer

And so the acceleration du to the external gravity to the mass m is not G. MG/(d2)
Proportional to the external mass MG and divided by the squared distance between m and MG

What is the correct acceleration in the relativistic context ?
 
  • #5
John SpaceY said:
TL;DR Summary: Link between speed and external gravity to a moving mass m (a spacecraft for example)

Hello,
I have a question linked to gravity and speed of light :

According to the Special Theory of Relativity, the speed of light cannot be exceeded because it would need an infinitive energy to accelerate the relativistic mass of a moving objet (a space craft for example) : indeed if the moving object has a mass m at rest, this mass would become M (the relativistic mass) and this mass M will tend towards infinity when the speed v of m will go towards c.
In the geometry of Special Relativity, there is no such thing as a relative speed greater than c. Light does not cause that geometry; it simply exists within that geometry, just like everything else. A special relationship that light (and all electromagnetic radiation) in a vacuum has with SR is that it is at the upper limit of relative velocity and therefore is strong evidence of the geometry.
 
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  • #6
John SpaceY said:
Thanks Dale for the answer

And so the acceleration du to the external gravity to the mass m is not G. MG/(d2)
Proportional to the external mass MG and divided by the squared distance between m and MG

What is the correct acceleration in the relativistic context ?
As relativistic speeds don't add as you expect by your daily experience, neither acceleration does to the speed... you can be accelerating forever, and you will not exceed the speed c for any observer. Acceleration acts hyperbolically and not parabolically, so to speak.
 
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  • #7
John SpaceY said:
What is the correct acceleration in the relativistic context ?
Zero. In relativistic gravity objects in free fall feel no force. The closest thing there is to an equivalent to the Newtonian concept of "acceleration due to gravity" is the acceleration required to hover at constant altitude, which clearly doesn't cause you to accelerate above light speed or we'd all be travelling faster than light just by standing here.
 
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  • #8
John SpaceY said:
What is the correct acceleration in the relativistic context ?
You have asked a trick question, probably without realizing it.

There are two possible answers.
The first is “zero”, because an accelerometer connected to an object moving under the influence of gravity will read zero. This quantity is called the proper acceleration, and is non-zero only for objects being forced off their natural inertial free-fall path by some external force.
The second is “whatever you want it to be”, if you mean the rate at which the speed of the object is changing. This is called coordinate acceleration, and we can make it come out to be whatever we please by choosing what we’re calculating the speed relative to.
 
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  • #9
Nugatory said:
You have asked a trick question, probably without realizing it.

There are two possible answers.
The first is “zero”, because an accelerometer connected to an object moving under the influence of gravity will read zero. This quantity is called the proper acceleration, and is non-zero only for objects being forced off their natural inertial free-fall path by some external force.
The second is “whatever you want it to be”, if you mean the rate at which the speed of the object is changing. This is called coordinate acceleration, and we can make it come out to be whatever we please by choosing what we’re calculating the speed relative to.
In fact I think that the answer is not "zero" but I don't know how to calculate in a relativistic context !
Suppose the spacecraft is going at 0,9c and very near to a "black-hole", the spacecraft will be attracted by the "black-hole" and the acceleration will no be "zero" : the speed of the spacecraft will continue to increase because of the acceleration of the external gravity and this acceleration is not linked to the mass of the spacecraft (and so not linked to its relativistic mass). If I am using Newtonian formulas the speed will increase c very soon but maybe I cannot use these formulas (as someone has written) and so I would like to know how to use the good formulas in order to know what will be the maximum speed of the spacecraft in this case (function of the distance between m and MG)
 
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  • #10
John SpaceY said:
In fact I think that the answer is not "zero"
If you are not going to accept the answers then why ask the question?

In fact, zero is the best answer by relativity. In relativity the acceleration is measured by an accelerometer. When an object is in free-fall its accelerometer reads zero. As @Nugatory there is a subtlety in that you could be asking about coordinate acceleration rather than the proper acceleration. But the "default" meaning of acceleration is the accelerometer acceleration which is 0.
 
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  • #11
John SpaceY said:
the spacecraft will be attracted by the "black-hole" and the acceleration will no be "zero" : the speed of the spacecraft will continue to increase because of the acceleration of the external gravity
If we choose to consider the black hole to be stationary, then yes, the coordinate acceleration of the spacecraft will be non-zero. But we could just as reasonably choose to consider the spacecraft to be at rest - and then its coordinate acceleration will be zero while the black hole's coordinate acceleration will be nonzero. Or we could choose some other object to be "at rest" and then both spacecraft and black hole would have non-zero coordinate acceleration.
(But be aware that using a black hole instead of, for example, a neutron star or other strongly gravitating massive object introduces some complications that don't belong in a B-level thread. There's nothing in your hypothetical that requires specifically a black hole so I am skipping over these subtleties).
 
  • #12
John SpaceY said:
In fact I think that the answer is not "zero" but I don't know how to calculate in a relativistic context !
What you are attempting to describe is coordinate acceleration. There is no single answer to what this is since it depends on your choice of coordinate system. That also tells you that it's not meaningful in of itself because no physics can depend on your choices. The only physically meaningful acceleration here is the proper acceleration, which is zero.
 
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  • #13
Ibix said:
What you are attempting to describe is coordinate acceleration. There is no single answer to what this is since it depends on your choice of coordinate system. That also tells you that it's not meaningful in of itself because no physics can depend on your choices. The only physically meaningful acceleration here is the proper acceleration, which is zero.
Thanks for the remarks
In fact I place myself on the surface of a black-hole and I see in the sky a spacecraft moving at 0,9c : the speed of the spacecraft of mass m is the speed I am seing from the surface of the blackhole.
And I see also that this mass m is very near to me and so there is an attraction force and a acceleration (not linked to the mass m). From where I am, I think that the attraction force will increase the speed of m and this speed will so be more than 0,9c.
I would like to calculate the speed, seen by me, function of the distance between m and the surface of the blackhole (the point where I am)
And the calculation has to be done in the relativistic context (with the good formulas ...)
 
  • #14
John SpaceY said:
In fact I place myself on the surface of a black-hole
You can't. You can hover above it, though.
John SpaceY said:
I see in the sky a spacecraft moving at 0,9c : the speed of the spacecraft of mass m is the speed I am seing from the surface of the blackhole.
It needs to be very near you for this to make sense, but ok.
John SpaceY said:
And I see also that this mass m is very near to me and so there is an attraction force and a acceleration (not linked to the mass m).
No, there is no force. The only force you feel is the acceleration of your rockets keeping you hovering. Gravity is not a force in general relativity and it does not accelerate you in the way you are trying to think.
John SpaceY said:
I would like to calculate the speed, seen by me, function of the distance between m and the surface of the blackhole (the point where I am)
And the calculation has to be done in the relativistic context (with the good formulas ...)
This is not well defined in curved spacetime so there is no unique answer. You can calculate the speed relative to local hovering observers if you like. The result will always be less than ##c##,
 
  • #15
Dale said:
But the "default" meaning of acceleration is the accelerometer acceleration which is 0.
Ok, but what if I have a "doppler accelerometer" (do this thing exist?), and I measure acceleration by the rate of change of the doppler shift from a distant star?
 
  • #16
Lluis Olle said:
Ok, but what if I have a "doppler accelerometer" (do this thing exist?), and I measure acceleration by the rate of change of the doppler shift from a distant star?
That wouldn’t change the default meaning at all. I am not sure why you think it would.
 
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  • #17
Lluis Olle said:
Ok, but what if I have a "doppler accelerometer" (do this thing exist?), and I measure acceleration by the rate of change of the doppler shift from a distant star?
That's a form of coordinate acceleration, because you need to define space and time so you can work out what "not kinematically Doppler shifted" looks like.
 
  • #18
Dale said:
That wouldn’t change the default meaning at all. I am not sure why you think it would.
Because you can measure something, which is different to nothing.
 
  • #19
Ibix said:
That's a form of coordinate acceleration, because you need to define space and time so you can work out what "not kinematically Doppler shifted" looks like.
Well, "my accelerometer" will look different away from the massive object, than when I'm near to the massive object of the OP. Seems that I can measure that something is happening.
 
  • #20
Lluis Olle said:
Well, "my accelerometer" will look different away from the massive object, than when I'm near to the massive object of the OP. Seems that I can measure that something is happening.
Sure. You just can't separate the effect of a changing path through curved spacetime from a change in your velocity without first defining what you mean by space and time - i.e., picking a coordinate system.
 
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  • #21
Lluis Olle said:
Ok, but what if I have a "doppler accelerometer" (do this thing exist?), and I measure acceleration by the rate of change of the doppler shift from a distant star?
Inferring relative coordinate velocity and acceleration from changes in redshift/blueshift only works in the special case of flat spacetime. For an example of why, consider what happens when either the source or the receiver slowly change their height in a gravity well.

Lluis Olle said:
Well, "my accelerometer" will look different away from the massive object, than when I'm near to the massive object of the OP. Seems that I can measure that something is happening.
Sure... You've discovered that things are different in different places.

(And takes off poster hat, puts on mentor hat you are coming very close to hijacking this thread. We have many many older threads explaining why the concept of relative velocity is so problematic in curved spacetime. Read some of them, and then if you have more questions start a new thread)
 
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  • #22
Lluis Olle said:
Well, "my accelerometer" will look different away from the massive object, than when I'm near to the massive object of the OP. Seems that I can measure that something is happening.
It's not a question of what you "measure" as acceleration, it is a question of what you (and an accelerometer) feel. Measurements without a felt acceleration are coordinate accelerations. We have to distinguish between real acceleration and coordinate acceleration all the time.

Consider an extreme example. Suppose you are a pilot in a jet going forward at 600 mph and your coordinate system is attached to your head, with the X-axis forward out your nose. The Y-axis is out your right ear. If you turn your head to the left, your X-velocity goes from 600 mph to 0 mph and your Y-velocity goes from 0 to 600 mph in one second. That is a huge coordinate acceleration. A "doppler accelerometer" attached to your head would measure huge numbers. But nothing physical has happened and you feel nothing.
 
  • #23
Lluis Olle said:
Because you can measure something, which is different to nothing.
Sure, but that wouldn’t change the default at all, so this little tangent is rather irrelevant.

Please try to support the OP by providing whatever insight you have to their question and avoid pointless digressions to others that do not aid either the OP or those who are trying to help the OP learn.
 
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  • #24
Thanks everybody for all your remarks

I think that I have understood my mistake : I need to change the newton parameters by Relativistics parameters in my equations.

I have found, using the Newton equations, that the speed of the spacecraft will be increased by dv when approaching a big mass MG (a black hole for example of mass MG) and this dv will be :

1676982681986.png


G is the Gravitational constant

MG is the mass of the external gravity to m

r is the distance between m and the center of MG

But in the Relativistic context, I have to change dr by the relativistic value of dr and this relativistic value will tend towards 0 when v will tend towards c

And so dv will tend towards 0 and the acceleration will tend towards 0 also when v will tend towards c. And at the end v will never be higher than c.
 
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  • #25
John SpaceY said:
But in the Relativistic context, I have to change dr by the relativistic value of dr and this relativistic value will tend towards 0 when v will tend towards c
No, that is not the relativistic law of gravitation. As I said in my first reply to get a relativistic law of gravitation you need to use the Einstein field equations.
https://en.wikipedia.org/wiki/Einstein_field_equations

These are fundamentally different from Newtonian gravity. You cannot simply substitute some "relativistic values" into Newtonian gravity and get GR.
 
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  • #26
John SpaceY said:
And at the end v will never be higher than c.
Your reasoning makes no sense because you are still trying to treat General Relativity as if it were Newtonian gravity with a few tweaks, rather than the reality that it is a completely different model with a completely different approach to most things. In particular, it doesn't model gravity as a force at all and you have no justification for your claim that ##dr## goes to zero as ##v## goes to ##c## (I suspect you've done some horrible twisting of length contraction, which doesn't apply here). However, ##v<c## is the correct conclusion.

The correct calculation yields that if the ship has initial velocity far from the black hole ##v_i## and you are hovering at radial coordinate ##r## above the black hole, the ship will pass you at ##c\sqrt{(c^2r_s-(r_s-r)v_i^2)/(c^2r)}##, where ##r_s## is the Schwarzschild radius. Note that the calculation used to get there is only valid for ##r>r_s##, and that the square root is less than 1 in that range. Note also that I'm being very careful to say that this is not a general expression for the speed of the ship (there isn't any such thing), but specifying that it's the speed measured at close range by a specific observer.
 
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  • #27
Thank you again for your answers Ibix and Dale !

Now I ask me another question which is linked to the last point : the rs value which is the Schwarzschild radius.

The time inside a moving object is reduced by the speed by a contracting factor which is

1676988534130.png

And we see here that when v tends towards c this coefficent tends towards 0 and so the time inside the moving object (a spacecraft for example) will tends towards 0

And also, when approaching an external gravity of mass MG the time inside a moving object is also reduced by another contracting factor which is

1676988548748.png

And we see here that when r tends towards rs this coefficent tends towards 0 and so the time inside the moving object (a spacecraft for example) will tends towards 0. The difference here is that r (the distance between the spacecraft of mass m and the center of the external gravity to m which has a mass MG) can have the value rs and so t can be equal to 0 and not only tends towards 0.

I can write the equality of the 2 contracting factors :

I arrive to

1676988572759.png

And I see here that when r = rs we have v = c

In other words, by going to rs, we have the same effect on the contraction of the time t inside the spacecraft than when v tends towards c. And also when r = rs we have v = c and not only v tends towards c.

And now I wanted to know what will happen if I continue to go at a distance lower than rs towards the center of MG, by a distance dr.

For this I have made a derivative of the function above and I have calculated dv/dr and the result is the following :

1676988623756.png

If I place myself at a distance lower than rs (difficult but theoretically possible for a black hole with a rs value of more than millions of km like Sagittarius A at the center of our galaxie), we see that dr will be negative and so dv will be positive !

And as I was already at v = c when r = rs, if I continue to go at a distance lower than rs, dv will be positive and the speed will be higher than c !

Is it correct for you ?
And now comes my new question :

When a spacecraft is travelling at a distance higher than rs, closed to a external gravity of mass MG (very high massn like a black hole for exmaple) and when the speed of the spacecraft is already near to c, by adding the contraction factor of the high gravity MG, could we consider to be in the same situation than when r = rs ?

And so when the spped is very near to c, at a distance r higher than rs, from the center of MG, could weconsider that this distance r could be the new rs ? and so, if the answer is yes, when we go at a distance lower than this r value, the speed v of the spacecraft will be more than c ! and this because we add the gravity effect to the speed effect and the speed effect was already at his maximum : and so adding the gravity effect at r will havethe same effect than adding the gravity effect at rs and so the speed v will exceed c at r, and so before rs !!!

Thanks in advance for your answers.
 
  • #28
John SpaceY said:
I can write the equality of the 2 contracting factors :
Why would you do that? This is not justified in any way that I can see.

John SpaceY said:
And I see here that when r = rs we have v = c
Which is a good indication that the calculation you are doing is nonsense. One issue that you are neglecting repeatedly is that (as I have said multiple times) you need to use the EFE. The EFE in turn uses the concepts of manifolds and coordinate charts. The most common coordinate chart for a spherical vacuum spacetime is the Schwarzschild chart. That chart is the one that has the ##\sqrt{1-r_s/r}## term you are looking at. However, that chart is undefined at ##r_s##, so you cannot use it to make any claims whatsoever about the behavior of physics at ##r_s##.

You really need to step back and learn to use the EFE and manifolds and coordinate charts before going further. I recommend Sean Carroll's lecture notes:
https://arxiv.org/abs/gr-qc/9712019

Please study and learn chapters 1-3 before proceeding with ANY more calculations. You can post questions about those chapters, but you must master that material before working on the things you are trying to work on here.
 
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  • #29
Ibix said:
Your reasoning makes no sense because you are still trying to treat General Relativity as if it were Newtonian gravity with a few tweaks, rather than the reality that it is a completely different model with a completely different approach to most things. In particular, it doesn't model gravity as a force at all and you have no justification for your claim that ##dr## goes to zero as ##v## goes to ##c## (I suspect you've done some horrible twisting of length contraction, which doesn't apply here). However, ##v<c## is the correct conclusion.

The correct calculation yields that if the ship has initial velocity far from the black hole ##v_i## and you are hovering at radial coordinate ##r## above the black hole, the ship will pass you at ##c\sqrt{(c^2r_s-(r_s-r)v_i^2)/(c^2r)}##, where ##r_s## is the Schwarzschild radius. Note that the calculation used to get there is only valid for ##r>r_s##, and that the square root is less than 1 in that range. Note also that I'm being very careful to say that this is not a general expression for the speed of the ship (there isn't any such thing), but specifying that it's the speed measured at close range by a specific observer.
Hello Ibix,
Thanks for the equation of the new v
But it is strange : I have made a numerical application and for example if I chose an initial speed vi of 0,9c and r = 10rs, I will find for the new speed v = 0,9c
The external gravity has not changed the initial speed and so ?
And it is the same for other numerical applications
Are you sure that nothing is missing in this equation ?
Thanks Ibix again
 
  • #30
John SpaceY said:
I will find for the new speed v = 0,9c
I get ##v=0.91 \ c##
 
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  • #31
Dale said:
I get ##v=0.91 \ c##
Yes sorry I have made an error in my calculation ...
I find now 0,91c
We see here that the speed of the spacecraft will increase thanks to the external gravity MG to m
And so, the acceleration is not zero ...
I agree that the acceleration will be zero when v will tend towards c and so the speed will never be higher than c but before reaching c the external gravity will increase the speed up to c.
Now the question is : if the spacecraft continu to go closer to MG, the speed is already to its maximum c, and what will be the effect of the external gravity in this case ? the external gravity reduces the time inside the spacecraft but the time is already at zero and so what is the next step ?
The equation before is only OK when r > rs but when r = 10 rs we are higher than rs and we are also higher than the radius of Sagittarius A for example (the radius of Sagittarius A is 12 millions of km and its rs value is around 3 millions of km).
 
  • #32
John SpaceY said:
And so, the acceleration is not zero
The coordinate acceleration is non-zero given the particular coordinate choice I made. It would be zero in other coordinate systems. The proper acceleration is zero independent of any choice I make. You will remain confused about what we are saying until you understand that there are two separate concepts here.
 
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  • #33
John SpaceY said:
And so, the acceleration is not zero ...
The coordinate acceleration is not 0 in the Schwarzschild coordinates. The proper acceleration is still 0, and the coordinate acceleration still must be specified relative to given coordinates.

John SpaceY said:
Now the question is : if the spacecraft continu to go closer to MG, the speed is already to its maximum c, and what will be the effect of the external gravity in this case ? the external gravity reduces the time inside the spacecraft but the time is already at zero and so what is the next step ?
The next step (and the previous several steps) is to learn chapters 1-3 of the lecture notes I posted for you earlier.

Please do not skip that this time.
 
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  • #34
Dale said:
The coordinate acceleration is not 0 in the Schwarzschild coordinates. The proper acceleration is still 0, and the coordinate acceleration still must be specified relative to given coordinates.The next step (and the previous several steps) is to learn chapters 1-3 of the lecture notes I posted for you earlier.

Please do not skip that this time.

Ibix said:
Your reasoning makes no sense because you are still trying to treat General Relativity as if it were Newtonian gravity with a few tweaks, rather than the reality that it is a completely different model with a completely different approach to most things. In particular, it doesn't model gravity as a force at all and you have no justification for your claim that ##dr## goes to zero as ##v## goes to ##c## (I suspect you've done some horrible twisting of length contraction, which doesn't apply here). However, ##v<c## is the correct conclusion.

The correct calculation yields that if the ship has initial velocity far from the black hole ##v_i## and you are hovering at radial coordinate ##r## above the black hole, the ship will pass you at ##c\sqrt{(c^2r_s-(r_s-r)v_i^2)/(c^2r)}##, where ##r_s## is the Schwarzschild radius. Note that the calculation used to get there is only valid for ##r>r_s##, and that the square root is less than 1 in that range. Note also that I'm being very careful to say that this is not a general expression for the speed of the ship (there isn't any such thing), but specifying that it's the speed measured at close range by a specific observer.
Hello Ibix,
I am French and I don't understand all the words and I have a question here :
when you write "you are hovering at radial coordinate r above the black hole" you want to say that the speed of the spacecraft is in the direction of the center of the black hole or the spacecraft is turning around the black hole at a distance r ?
For me I understand that the initial speed vi is in the direction of the center of the black hole and the new speed v is also in the same direction.

If the initial speed vi is in the direction of the center of the black hole and the new speed v is tangential to the black hole (the space craft is turning around the black hole) there is a change in the direction and I would like to understand why there is this change in the direction ? and what force has created this ?
 
  • #35
And also the equation is only OK for r > rs
Do you know what will happen for r < rs when the speed is near to c when r will arrive to rs ?
I understand it is not possible to go to rs, because there is some matter at r = rs, but it is for a theoretical calculation
Thanks in advance
 

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