I Is Absolute Position Necessary to Understand Relativity?

  • I
  • Thread starter Thread starter joshk
  • Start date Start date
  • Tags Tags
    Absolute
joshk
Messages
3
Reaction score
0
I am guessing this is an easy one to grasp, but I think I am missing something in my understanding of relativity.
Relativity suggests that as an object moves toward the speed of light, a greater amount of force is required to increase its velocity.
For this to be true, wouldn't it require the existence of absolute position?
For instance, when we say an object is approaching the speed of light, shouldn't this be relative to another inertial frame, such as an absolute inertial frame?
An apple is moving through space near the speed of light relative to the earth. If relativity is saying that a larger force (than predicted by Newtonian mechanics) is required to increase the velocity of the apple, then why wouldn't it be also be the case for a pear on Earth (who is also moving relative to the apple near the speed of light)?
 
Physics news on Phys.org
joshk said:
For this to be true, wouldn't it require the existence of absolute position?

No, because speed and force are frame-dependent.

joshk said:
when we say an object is approaching the speed of light, shouldn't this be relative to another inertial frame

It's relative to some inertial frame, yes. But there will always be some other inertial frame in which the object is at rest. Speed is frame-dependent.

joshk said:
such as an absolute inertial frame?

There is no such thing.

joshk said:
An apple is moving through space near the speed of light relative to the earth. If relativity is saying that a larger force (than predicted by Newtonian mechanics) is required to increase the velocity of the apple, then why wouldn't it be also be the case for a pear on Earth (who is also moving relative to the apple near the speed of light)?

The force an observer on Earth would have to exert on the apple would be larger. But the force someone moving along with the apple would have to exert would not. And conversely, an observer on Earth would not have to exert more force on the pear, but an observer moving along with the apple would. Speed and force are frame-dependent.
 
  • Like
Likes joshk
joshk said:
For instance, when we say an object is approaching the speed of light, shouldn't this be relative to another inertial frame
Yes it should. Also "a greater amount of force is required to increase its velocity" should read "a greater amount of force as measured in the frame in which the object is approaching the speed of light is required to increase its velocity".
joshk said:
If relativity is saying that a larger force (than predicted by Newtonian mechanics) is required to increase the velocity of the apple, then why wouldn't it be also be the case for a pear on Earth (who is also moving relative to the apple near the speed of light)?
Force is frame dependant. According to the pear's frame it requires more force to achieve the same coordinate acceleration of the apple as for the pear. According to the apple's frame, vice versa.
 
  • Like
Likes joshk
thanks peter and ibix that clears up a lot!
I did suspect that would be the explanation for the last question. So if I could follow up, I get confused in that...
If the force to increase the speed of the apple as you move along with the apple is Newtonian, then wouldn't it be possible to actually get the velocity of the apple relative to Earth to the speed of light as long as someone (or something like a rocket) is traveling along with and pushing the apple?
 
joshk said:
thanks peter and ibix that clears up a lot!
I did suspect that would be the explanation for the last question. So if I could follow up, I get confused in that...
If the force to increase the speed of the apple as you move along with the apple is Newtonian, then wouldn't it be possible to actually get the velocity of the apple relative to Earth to the speed of light as long as someone (or something like a rocket) is traveling along with and pushing the apple?
The problem is that someone in the spaceship measures time differently than someone on the Earth. So for example, if the ship is moving at 0.9c, it is perfectly possible for something in the rocket to be accelerated to 0.1c relative the the rocket as measured by someone in the rocket. However, someone on the Earth would only measure the object as being accelerated to 0.91743...c relative to the Earth.( and the object would only measure its velocity as being 0.91743...c relative to the Earth.)
 
  • Like
Likes joshk
Janus said:
The problem is that someone in the spaceship measures time differently than someone on the Earth. So for example, if the ship is moving at 0.9c, it is perfectly possible for something in the rocket to be accelerated to 0.1c relative the the rocket as measured by someone in the rocket. However, someone on the Earth would only measure the object as being accelerated to 0.91743...c relative to the Earth.( and the object would only measure its velocity as being 0.91743...c relative to the Earth.)

thanks that makes perfect sense!
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top