In e=mc^2 , ‘c’ really just equals 1 (unless you have engineering calcs to do). Can you apply the Lorentz Transformation to m=e by using metric tensor(flat spacetime) squared over metric tensor(curved spacetime) squared? Would that have any effect on our ability to estimate mass?
You can do all relativistic calculations in any set of units you want, as long as you remember to interpret the answers to be in the same units as you started with. However, it’s important to remember that if you’re using a c=1 unit system, that E still has the dimensions of mass times velocity squared, while m has the dimension of mass, so even when c=1 the 1 has the dimensions of velocity squared.
Standing by for Chronos and the other physics experts to elaborate on subtleties I’ve overlooked (or to correct flat-out errors I’ve made).
The c[sup]2[/sup] term is there because of the dimension of time that you can’t control.
The mass–energy equivalence cannot drop it in the combined space-time.
To expand on this, lets rewrite E = m[sub]0[/sub]c[sup]2[/sup] for mass.
The full form to solve for mass:
m[sub]0[/sub][sup]2[/sup] = (E/c[sup]2[/sup])[sup]2[/sup] - (p/c)[sup]2[/sup]
As p = 0 the momentum term vanishes.
m[sub]0[/sub][sup]2[/sup] = (E/c[sup]2[/sup])[sup]2[/sup]
Which simplifies to:
m[sub]0[/sub] = E/c[sup]2[/sup]
For objects that have no mass it reduces to:
p = E/c
Remember that things traveling at the speed of light do not experience time, and their creation and absorption is a single event.
While the formula does reduce to E = m[sub]0[/sub]c[sup]2[/sup] for a massive object in a center-of-momentum frame, that object still experiences time.
If a massfull product has any momentum you have to go back to the full form:
m[sub]0[/sub][sup]2[/sup] = (E/c[sup]2[/sup])[sup]2[/sup] - (p/c)[sup]2[/sup]
The mistaken concept of “relativistic” mass is contained in this form, so it useful to but if you re-write it for Energy:
If you play around with this form you will find that c[sup]2[/sup] term is what becomes your speed limit. Note I have used m[sub]0[/sub] here to make it very clear that this mass is the rest mass or the invariant mass of the system to avoid the typical relativistic mass pitfall. Many basic books still use relativistic mass, but just discard the concept in your own learning outside of those contexts and it will help.
I’m really unclear on what you mean by this, and I do this stuff for a living. In general, dividing a tensor by another tensor is not a well-defined mathematical operation. Can you elaborate on your question a bit?
Analogy to classic physics if the path time the formula where s looks like time and where the g[sub]ij[/sub] (metric tensor) is first seen.
Where u is a unit vector.
Note that this means if the speed of light is your unit we are always moving through space-time at the same speed. If we don’t move through space we have to move through time and if we go faster in space we go slower through time.
You can get to E=m in some reductive cases, but not all. As an example if you are considering the kinetic energy E=mc[sup]2[/sup] form allows for things like the atomic bomb.
I’m with MikeS in not understanding what the heck is going on here. In addition to the weirdness of the proposed calculations, I’m not sure how a different way of defining mass would aid in estimations of mass. What are we estimating the mass of?
I made the assumption that the OP was making the same mistake most people do when considering the implications of the equivalence of mass and energy as a general principle and consequence of the symmetries of space and time.
The way relativity is often introduced it is easy to miss that you have no prior geometry and the typical empty space:
x[sub]1[/sub],x[sub]2[/sub],x[sub]3[/sub],x[sub]4[/sub] → ds (proper time separation) → tangent vector (velocity) → geodesics (motion) → Riemann (flatness) → Ricci (gravity)
Hides the functions of establishing the geometry in the typical way it is explained.
As the Metric coefficients g[sub]ij[/sub] portion of ds[sup]2[/sup]=g[sub]ij[/sub]dx[sub]i[/sub]dx[sub]j[/sub] and Ricci tensor R[sub]ij[/sub]=0 in the special relativity case in a COM frame in empty space it is not an uncommon deduction.
Especially as natural units hiding the travel in the time dimension. All of the canceling out parts can make it seem like the Metric Tensor acts a bit differently then it does until you deal with the more realistic cases.
(yes I did the formulas in 2d, i’m lazy)
A year ago I asked if c could change near a black hole. There were some very patient folks who showed me why that question doesn’t make sense. So now my question is can the relationship between mass and energy change in extremely bent spacetime. I’m still reading the wiki article on tensors in curvilinear coordinates…
I know this is probably another stupid question.
With that information, and in over simplified terms, the metric tensor is what lets you measure the “distance” between events.
Two observers can not assume that the actual shape of space-time is the same, and actually they can’t assume the shape of their own space.
The metric tensor is a tool you look for to figure out the proper time interval Δτ between two events in your own reference frame. If you cannot find a metric tensor that works for your reference frame or a chosen reference frame you are blocked. Without finding it you simply cannot find a measurement to share with someone else that they will agree on.
Note again: This is a over simplification.
Remember that Eisenstein derived E=mc[sup]2[/sup] from Newton’s and laws with the assumption that time is consistent. The implications of c[sup]2[/sup] when solving for mass is that your cat contains enough energy to power most countries for an entire year if we could use it.
Most of what makes math look scary in general relativity is trying to figure out the shape of spacetime for the local observer.
Exact solutions for the EFE can only be found under simplifying assumptions such as symmetry. Special classes of exact solutions are most often studied as they model many gravitational phenomena, such as rotating black holes and the expanding universe. Further simplification is achieved in approximating the actual spacetime as flat spacetime with a small deviation, leading to the linearized EFE. These equations are used to study phenomena such as gravitational waves.
Is spacetime curvature independent of frame of reference, like the speed of a photon?
The curvature itself is invariant, but how it’s described might depend on reference frame. The most complete description of curvature is the Riemann tensor, which has 20 distinct components. The values of those individual components will vary with frame, just like the components of a vector might vary, even though the vector (or tensor) as a whole is the same. But there are various ways to summarize the Riemann tensor with a single scalar, such as the Ricci scalar, which are invariant.
Would a spacecraft traveling at .99c measure the deflection of a photon around a star the same as a spacecraft traveling at .0001c?
If you define the deflection in an invariant way, yes. People who do actual work in relativity do almost nothing with frame-dependant variables: All of the real work is done in invariant variables of various sorts. At most, you might get one conversion to frame-dependent variables for your final answer.
Another way to think about this is that angular momentum is the conserved quantity in this case. When a photon is “deflected” by a star it is actually not accelerated in any way. The photon is following the geodesic which exactly the same thing as a straight line in a curved space-time.
The photon is not actually being deflected at all, it is traveling in a straight line at a consent speed with no change to it’s momentum. What we view deflection is merely an artifact of us assuming that space-time is flat.
Noether’s Theorem is the subject that will help you with what the conserved quantities are BTW.
Erg… Pull angular off of momentum in that post…while angular momentum is conserved in GR, in this case it is just momentum.
To add some some basic “intuitions” on why if curvature is invariant yet more tests are required past figuring out the Riemannian metric.
The important thing to remember is that the surface may be flat but not Euclidean.
The Ricci part can be though of “magnification” or an object increasing in apparent volume while traveling on a geodesic. Consider a rain drop on your computer screen.
The Weyl part can be thought of as causing a circle to be squeezed look like a football or while traveling on a geodesic.
These are just analogies and intuitions, but it is helpful that just because a manifold is flat doesn’t imply it it is euclidean.
In most problems these zero out, but they should hopefully help illustrate that you can find useful intuitions for these concepts and that you have to drop the Euclidean assumptions when working in this space until you know that you can make those assumptions again.
It helps if you focus on conserved quantities and symmetries and not on frame dependent observations.
If the relationship between mass and energy did change in extremely bent spacetime, would we already know that? Would it make any difference to our current understanding of physics?
Modifications to the theory of gravitation would make a huge difference to our understanding of physics. Also to astronomy, etc.
Remember that mass is just a dense form of energy and both are interchangable.
That means any form of energy can curve the space. The time–time component of the of stress-energy tensor is the density of the M in E=mc[sup]2[/sup].
Mass is just the energy density divided by the speed of light squared. It is just one of 16 components in the stress-energy tensor. The gravitational field can accomplish work with matter; and matter can accomplish work with the gravitational field so the conserved properties change.
Mass is energy, and the form of energy changes all of the time.
Most likely a future experiment will find some form of Lorentz invariance violation which will lead to new physics. But due to the massive amounts of very accurately predicted experimental data we have it is also highly unlikely that this newer physics will be dramatically different than GR. But who knows and I just hope to live long enough to hear about it.
The hard distinction you are making between matter and energy just doesn’t work in this domain.