Relativity when you divide a trip into small time intervals

Click For Summary
SUMMARY

This discussion centers on the implications of Planck length and Planck time in the context of relativistic motion. The user posits that an object moving at less than the speed of light (C) spends time stationary, leading to the conclusion that any observed speed less than C is an average based on periods of motion at C and periods of being stationary. The conversation highlights the complexity of applying discrete intervals to continuous relativistic effects, ultimately questioning the compatibility of relativity with a discretized model of space-time.

PREREQUISITES
  • Understanding of Planck length and Planck time
  • Familiarity with the concept of speed of light (C)
  • Basic knowledge of four-velocity in relativity
  • Awareness of relativistic effects and time dilation
NEXT STEPS
  • Research the implications of Planck units in quantum gravity
  • Study the concept of four-velocity and its mathematical representation
  • Explore the relationship between speed, time dilation, and relativistic effects
  • Investigate the challenges of discretizing space-time in physics
USEFUL FOR

Physics students, researchers in theoretical physics, and anyone interested in the foundations of relativity and quantum mechanics.

1MileCrash
Messages
1,338
Reaction score
41
Disclaimer: I am still learning physics. I think about weird things. This results in thoughts like the one that follows.



My understanding of Planck length is that it is the smallest increment of length that can be measured, and to me this implies that an object can only be a whole number of Planck lengths away from another.

Planck time, as I understand it, is the smallest possible increment of time in which any "event" can actually occur. Since it is derived from the time it takes moving at the fastest possible speed across the smallest possible distance, this makes perfect sense.

This also means that 1 Planck length per 1 Planck time = motion at the speed of light.

Does this imply that when any object moves at a speed less than C, it is really spending a fraction of this motion completely stationary?

If I watch a ball moving at 1/2 the speed of light, noting how many Planck lengths it moves per Planck time, the only result that makes sense with the above given is that during half of the Planck time increments, it moves one Planck length, and for the other half of time intervals, it doesn't move at all.

Since I couldn't observe it moving any fraction of a Planck-length, and I couldn't observe anything happen in a fraction of a Planck time, my only conclusion is that in this *thought experiement* is that I would observe it half the time as moving at the speed of light, and half the time not moving at all. In other words, I couldn't observe it moving one-half of a Planck length per Planck time.

Thus, to make my title clear, this means to me, that if something is actually moving, it is moving at C. Any speed witnessed less than C is an average speed based on how much time is spent stationary or spent moving at C - the only two possibilities.

But here's where I get a bit confused.

If an object is traveling at the 50% lightspeed relative to me, and I am Chuck Norris, I should be able to divide the journey into Planck time intervals and see that half the time, it is stationary, and half the time, it is moving at light speed. However, this would also suggest that half the time, it experiences no elapsed time relative to me, and half the time, it experiences the same amount of elapsed time relative to me, suggesting that in total, it experiences half of the elapsed time relative to me. This is of course not the case, as time only elapses 1.154 times quicker at that speed for the traveler relative to me.

So why is it when I divide the trip this way, the relativistic effects don't make sense?
 
Physics news on Phys.org
Relativity really isn't meant to "work" with discretization of the underlying space. The problems with doing so are why a theory of quantum gravity elude us so.

Nevertheless, you're correct to say that any object's velocity is c at all times. This exactly gets at the idea of the four-velocity, which is usually denoted u. An object traveling with ordinary (or "three-") velocity v = \beta c in the e_x direction has four-velocity

u = \gamma c (e_t + \beta e_x)

When you take the magnitude of this vector, you always get c, regardless of \beta (note that \gamma = (1-\beta^2)^{-1/2}, which is why the normalization stays fixed).
 
1MileCrash said:
My understanding of Planck length is that it is the smallest increment of length that can be measured, and to me this implies that an object can only be a whole number of Planck lengths away from another.

That's not true, so the rest of what you wrote doesn't follow.
 

Similar threads

  • · Replies 40 ·
2
Replies
40
Views
7K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 43 ·
2
Replies
43
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 9 ·
Replies
9
Views
935
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K