Theory of Relativity, "you can't tell which mass is moving"

[TheScienceOrca]

Hello, I thought of something after watching this video on the theory of relativity.


It says "it is impossible to tell which object is moving"

Well I have now thought of a scenario where it would be possible.

We can observe and measure the delta in speeds of the two masses as they pass by but that won't tell us which one is moving.


But I have thought of special cases where you WOULD be able to tell which seems strange.


If you were on a rocket in an imaginary universe going .5c + 1m/s and another mass was going .5c + 1m/s towards you, you could measure the delta speeds as c + 2 m/s

Since no mass can travel faster than the speed of light. Both masses most be moving.


Please share your feedback

 

[ghwellsjr]

Hello, I thought of something after watching this video on the theory of relativity.


It says "it is impossible to tell which object is moving"

Well I have now thought of a scenario where it would be possible.

We can observe and measure the delta in speeds of the two masses as they pass by but that won't tell us which one is moving.


But I have thought of special cases where you WOULD be able to tell which seems strange.


If you were on a rocket in an imaginary universe going .5c + 1m/s and another mass was going .5c + 1m/s towards you, you could measure the delta speeds as c + 2 m/s

Since no mass can travel faster than the speed of light. Both masses most be moving.


Please share your feedback

You are right, no mass can travel faster than the speed of light. But you didn't say either mass was traveling faster than the speed of light. If I understand you correctly, you are saying that I am traveling at just over half the speed of light in one direction and a mass is coming towards me at the same speed, that would mean that it was stationary according to the frame in which I was moving, correct?

When you speak of the speed of an object, it is always according to a reference frame. In that reference frame, no mass can travel faster than the speed of light. But you can always transform to another reference frame moving at any speed short of c and in any direction with respect to the first frame and then all the masses could be traveling at different speeds including zero. That is what is meant by the expression, "you can't tell which mass is moving", because all reference frames are equally valid.

Does that make sense to you?

 

[TheScienceOrca]

You are right, no mass can travel faster than the speed of light. But you didn't say either mass was traveling faster than the speed of light. If I understand you correctly, you are saying that I am traveling at just over half the speed of light in one direction and a mass is coming towards me at the same speed, that would mean that it was stationary according to the frame in which I was moving, correct?

When you speak of the speed of an object, it is always according to a reference frame. In that reference frame, no mass can travel faster than the speed of light. But you can always transform to another reference frame moving at any speed short of c and in any direction with respect to the first frame and then all the masses could be traveling at different speeds including zero. That is what is meant by the expression, "you can't tell which mass is moving", because all reference frames are equally valid.

Does that make sense to you?

Ok, let's go from another observer. You could still see two objects traveling in towards each other at .5c + 1m/s

This means that both are moving, not one of those objects can be stationary.

Also keep in mind no masses are traveling at the speed of light, they are traveling .5c + 1m/s


Think of driving town a 2 lane road at 1 M/s in the opposite lane a car approaches at 1m/s

The delta is 2m/s even though no object is going 2M/s

 

[Orodruin]

In *your* reference frame both are moving. This does not mean that they can tell which is moving (or that your assertion that they are moving is correct - they will consider you to be moving). The difference of velocities depends on the frame. In the case of linear motion in one dimension, there is an easy expression for adding velocities, in your case it reduces to:
$$
v' = \frac{2\cdot (0.5 c + 1\,{\rm m/s})}{1 + \frac{(0.5c + 1\,{\rm m/s})^2}{c^2}}
\simeq
c\frac{1 + 3\cdot 10^{-9}}{1 + 0.25 + 3\cdot 10^{-9}} \simeq 0.8c.
$$
Thus, in the rest frame of one of the ships, the other ship will seem to be moving at about $0.8c$.

 

[TheScienceOrca]

In *your* reference frame both are moving. This does not mean that they can tell which is moving (or that your assertion that they are moving is correct - they will consider you to be moving). The difference of velocities depends on the frame. In the case of linear motion in one dimension, there is an easy expression for adding velocities, in your case it reduces to:
$$
v' = \frac{2\cdot (0.5 c + 1\,{\rm m/s})}{1 + \frac{(0.5c + 1\,{\rm m/s})^2}{c^2}}
\simeq
c\frac{1 + 3\cdot 10^{-9}}{1 + 0.25 + 3\cdot 10^{-9}} \simeq 0.8c.
$$
Thus, in the rest frame of one of the ships, the other ship will seem to be moving at about $0.8c$.

Thank you for the speedy reply and doing the equations for me I appreciate it!

Ok let's add another ship would that take the delta over c if you were standing on a planet looking at the 3 rockets going .9c you could measure that all objects must be moving because the total delta would be over 2c.

2c is the max possible relative motion if 1 object is stationary.

 

[ghwellsjr]

Thus, in the rest frame of one of the ships, the other ship will seem to be moving at about $0.8c$.

Better to say the other ship is moving at about $0.8c$. We don't want anyone to think that it's just an illusion.

 

[ghwellsjr]

Thank you for the speedy reply and doing the equations for me I appreciate it!

Ok let's add another ship would that take the delta over c if you were standing on a planet looking at the 3 rockets going .9c you could measure that all objects must be moving because the total delta would be over 2c.

2c is the max possible relative motion if 1 object is stationary.

Did you actually do the calculations?

 

[TheScienceOrca]

Did you actually do the calculations?

I didn't do the relative motion calculation above, because I am unaware of the syntax for that equation.So if there are 3 objects in question, here is my logic.

I know no object can travel greater than c.

Lets say in extreme scenario the max possible relative motion if one of the 3 objects is stationary is 2c.

How did I get this?I got this because 2 objects traveling at c = 2c.Now if there were 3 objects traveling at .9c the relative motion would be greater than 2c which means no object can be at rest.

 

[ghwellsjr]

I didn't do the relative motion calculation you did above, because I am unaware of the syntax for that equation.


So if there are 3 objects in question, here is my logic.

I know no object can travel greater than c.

Lets say in extreme scenario the max possible relative motion if one of the 3 objects is stationary is 2c.

How did I get this?


I got this because 2 objects traveling at c = 2c.


Now if there were 3 objects traveling at .9c the relative motion would be greater than 2c which means no object can be at rest.

You need to state your scenario clearly and precisely. Let's say that you have a frame of reference in which the first object is traveling at 0.9c. Then let's say that relative to the first object, a second object is traveling in the same direction at 0.9c relative to the first. Finally, let's say that a third object is traveling in the same direction at 0.9c relative to the second object. Now we ask the question, how fast is the third object traveling according to our frame of reference? You might think the answer should be 2.7c, faster than the speed of light.

But according to the formula, the second object is traveling at 0.994475c in the reference frame. Now we apply the formula again for the third object and get 0.9997c. We keep getting closer to c but never arriving at c.

 

[TheScienceOrca]

You need to state your scenario clearly and precisely. Let's say that you have a frame of reference in which the first object is traveling at 0.9c. Then let's say that relative to the first object, a second object is traveling in the same direction at 0.9c relative to the first. Finally, let's say that a third object is traveling in the same direction at 0.9c relative to the second object. Now we ask the question, how fast is the third object traveling according to our frame of reference? You might think the answer should be 2.7c, faster than the speed of light.

But according to the formula, the second object is traveling at 0.994475c in the reference frame. Now we apply the formula again for the third object and get 0.9997c. We keep getting closer to c but never arriving at c.

I see thank you so much for the help and getting that off my mind.

I appreciate it!

 

[ghwellsjr]

I see thank you so much for the help and getting that off my mind.

I appreciate it!

You're very welcome.