# GR gravity path confusion

• B
stevendaryl
Staff Emeritus
Science Advisor
You don't know that!
Okay, all the evidence points to a Big Bang cosmology that does have a unique directionality in time. The model could be wrong, of course, but just about any non-tautological statement could be wrong.

nnunn
Ok. I get that objects seem to fall because of curved spacetime, when they're actually just moving in straight paths. I get the example of ants walking in straight lines on the surface of a sphere, thinking that something attracts them to each other. What I don't get is how the "ants" are "attracted" to each other without walking. What that means is, how is a stationary object attracted to another stationary object? I don't get this.
Remember that 4D spacetime is curved, not 3D space. A stationary object in spacetime is a line, not a point. A stationary object in space is moving into the future.

And the point is that the spacetime path of any particle will have a nonzero "velocity" with respect to any timelike axis.
As long as the particle has mass.

PeterDonis
Mentor
2020 Award
the spacetime path of any particle will have a nonzero "velocity" with respect to any timelike axis.
Which mathematical object are you referring to here? The norm of the tangent vector to the particle's worldline? If so, MeJennifer's statement that this only applies to a massive particle is correct. However, the norm is invariant and you seem to be talking about something that could change from one inertial frame (choice of timelike axis) to another, so I'm not sure exactly what mathematical object you are referring to.

stevendaryl
Staff Emeritus
Science Advisor
Which mathematical object are you referring to here?
The component of the 4-velocity of the particle in the "time" direction. Mathematically, let $e_\mu$ be any basis for which $e_0$ is timelike and $e_j$ is spacelike, for $j \neq 0$. Let $U^\mu$ be a particle's 4-velocity expressed in this basis. Then $U^0 \neq 0$. I think that's true. If so, it's a statement that is true in any locally Minkowskian frame, but it's not clear how to express it as a statement about invariants.

vanhees71
Science Advisor
Gold Member
A classical particle is always on-shell and thus you have
$$g_{\mu \nu} u^{\mu} u^{\nu}=1$$
since
$$u^{\mu}=\frac{1}{m} p^{\mu},$$
where ##p^{\mu}## is the momentum of the particle.

The four-velocity is always time like and ##u^0 >0## by convention, i.e., you take the proper time defining the same causal time direction as the coordinate time.

stevendaryl
Staff Emeritus
Science Advisor
The four-velocity is always time like and ##u^0 >0## by convention, i.e., you take the proper time defining the same causal time direction as the coordinate time.
$u^0 \neq 0$ is what I was trying to say in post #25.

A mass less particle does not have a four velocity.

stevendaryl
Staff Emeritus
Science Advisor
A mass less particle does not have a four velocity.
It has a parametrized path $x^\mu(s)$ (although $s$ is not unique), and associated with every parametrized path is a 4-vector whose components are $\frac{d x^\mu}{ds}$. The only difference with a massive particle is that $s$ can't be proper time.

vanhees71
Science Advisor
Gold Member
A mass less particle does not have a four velocity.
True, but there are no massless particles (no, photons are no particles!).

True, but there are no massless particles (no, photons are no particles!).
Whether photons are particles or not, which is more like a definition question anyway, seems to me not directly relevant to general and special relativity.

vanhees71
Science Advisor
Gold Member
Well, it depends on what you are interested in. Photons are absolutely crucial in quantum optics. If you are interested on classical physics only, you don't need them, and it's always much better to stick to classical descriptions, i.e., the Maxwell equations in this case.

PeterDonis
Mentor
2020 Award
The component of the 4-velocity of the particle in the "time" direction.
I would use the term "tangent vector" here in order to avoid the complaints you are getting that a massless particle does not have a 4-velocity (which is supposed to be a unit vector). With that correction, I agree that the "0" (timelike) component of the tangent vector to any timelike or null curve will be nonzero with respect to any orthonormal basis.

Dale
PeterDonis
Mentor
2020 Award
there are no massless particles (no, photons are no particles!).
This is a "B" level thread, so all of the issues you are implicitly referring to here are really beyond the scope of this discussion. For purposes of a "B" level thread, I think the term "massless particles" is an acceptable way to refer to photons, as long as everyone is aware that it's a "B" level heuristic.

If you are interested on classical physics only, you don't need them
You don't "need" them in the sense that you can always generate predictions without using the concept, yes. But for this kind of discussion, where we can use the extreme geometric optics approximation and don't really care about the details of Maxwell's Equations, thinking of photons as "massless particles"--pointlike objects that move on null worldliness--can be very useful to avoid cluttering up your descriptions with unnecessary details.

You're thinking of curved space only. In curved spacetime, stationary objects are effectively moving along the time axis at speed c, so the curvature of free fall paths as the object moves through time accelerates it even if it isn't moving in space.
I had understood that we were moving through time fast, but at c is very fast. Are you sure? Can you show a mathematical proof that is fairly simple?

If Einstein knew that the movement through time caused gravity, then why did he say that time is an illusion?

PeterDonis
Mentor
2020 Award
If Einstein knew that the movement through time caused gravity, then why did he say that time is an illusion?
Even though this is a "B" level thread, rules about acceptable sources still apply. Einstein didn't say that time is an illusion in any scientific paper that I'm aware of. He may have said it in pop science books or articles, but those aren't acceptable sources for discussion here.

Also, "movement through time causes gravity" is not a good way of describing what has been said in this thread. The idea that all objects are "moving through time" along their worldlines in a curved spacetime can help to understand how the motion of those objects shows the effects of gravity; but that's not the same as saying the "movement through time" of those objects causes gravity.

It has a parametrized path xμ(s)xμ(s)x^\mu(s) (although sss is not unique), and associated with every parametrized path is a 4-vector whose components are dxμdsdxμds\frac{d x^\mu}{ds}. The only difference with a massive particle is that sss can't be proper time.
Should that not read "massless" instead of "massive" ?

stevendaryl
Staff Emeritus
Science Advisor
Should that not read "massless" instead of "massive" ?
I meant the only difference between a massless and massive particle is that a massless particle's path cannot be parametrized by proper time.

Markus Hanke