How does the weight of a rock change in relativistic speeds?

  • Thread starter The Rev
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In summary, Person A is traveling at relativistic speeds and holding a stone over his head. From Person A's point of view, the rock is unchanged, so he continues to hold it up. From Person B's point of view, however, the rock grows heavier due to relativistic increases in mass, and falls on Person A's head, killing him. To communicate to Person B that he is still alive, Person A would have to use some form of force that is not affected by relativistic time dilation.
  • #1
The Rev
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I'm pretty new to the ideas of relativity, so maybe this is old ground to cover for some of you. However, I need to know how to resolve an apparent paradox.

Person A is traveling at relativistic speeds holding a stone over his head. The stone weighs just slightly less than the maximum weight that he is able to hold up. Person B is an observer, watching Person A. From Person A's point of view, the rock is unchanged, so he continues to hold it up. From Person B's point of view, however, the rock grows heavier due to relativistic increases in mass, and falls on Person A's head, killing him.

Now, how does Person A, who is alive from his own point of view, communicate to Person B that he really isn't dead?

o:)

The Rev
 
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  • #2
SR doesn't deal with gravity and unfortunately relativstic descriptions of grabity can never be that simple. Also remember F=Ma isn't generally true in SR anyway.
 
  • #3
It turns out the rock never gets heavier in SR. It either has the same weight, or it gets lighter! (By "weight" I mean "force").

It's the same weight (force!) if the force points in the same direction as the observer is moving.

It's lighter (a lower force) if the force points at at an angle to the direction of motion.

This comes from the transformation laws, which can be derived from the folowing defintions.

4-force = dp/dtau, where p is energy-momentum 4-vector, and tau is the "proper time".

4-forces transform like any other 4-vector.

3-force = dp/dt - where p is the relativistic momentum (the energy-momentum 4-vector with the energy part omitted), and t is the coordinate time, rather than the proper time.

From the above, it can be seen that the 4-force and the 3-force are related by the time dilation factor, dt/dtau, i.e. (dp/dtau) = (dp/dt) * (dt/dtau)
 
  • #4
pervect said:
It turns out the rock never gets heavier in SR. It either has the same weight, or it gets lighter! (By "weight" I mean "force").

It's the same weight (force!) if the force points in the same direction as the observer is moving.

It's lighter (a lower force) if the force points at at an angle to the direction of motion.
Just to avoid confusion, the weight cannot get lighter in the frame of the man carrying the boulder, since that would provide the man with a way of measuring his absolute velocity...I guess the resolution would be that even if the force from the boulder pushing down on the man weakens as seen in some other frame, the force upwards from the man pushing on the boulder will also weaken as seen in that frame, so this frame will also predict no change in how high the boulder is held up.
 
  • #5
Jscd : This is an interesting problem you put forward :

1)in SR, time and space change depend on the relative speed
2)in GR time and space depend on the relative position towards the source of gravitation.

shouldn't in GR space-time depends on the relative motion of the observer towards the source of gravitation too ?
 
  • #6
For the guy hodling the rock, the weight he must hold doesn't change because his
time is slowing down. A heavier rock in slower time doesn't weigh more because there
is a time derivative in computing the force.
 
  • #7
For the guy hodling the rock, the weight he must hold doesn't change because his
time is slowing down. A heavier rock in slower time doesn't weigh more because there
is a time derivative in computing the force.


just for the record, what force are you talking about, and in regards to what time derivative?
 

Related to How does the weight of a rock change in relativistic speeds?

1. What is a simple SR problem?

A simple SR problem refers to a problem in which the source and receiver of a signal are in close proximity to each other and there is minimal interference or distortion in the signal due to distance.

2. How is a simple SR problem different from a complex SR problem?

A simple SR problem is different from a complex SR problem in that a complex SR problem involves more factors that can affect the signal, such as distance, obstacles, and interference from other sources. In a simple SR problem, these factors are minimal.

3. What are the common causes of a simple SR problem?

The most common cause of a simple SR problem is when the source and receiver are too close to each other, resulting in a strong and clear signal. Other causes may include a lack of interference or obstacles between the source and receiver.

4. How can a simple SR problem be solved?

A simple SR problem can be solved by adjusting the distance between the source and receiver, using shielding or filters to minimize interference, or by using a more powerful source or sensitive receiver.

5. Why is understanding simple SR problems important in scientific research?

Understanding simple SR problems is important in scientific research because it allows for accurate and reliable data collection and analysis. It also helps to identify potential sources of error and improve the quality of research results.

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