The physics department of our school does not have a course related to the theory of relativity

First, basic courses such as advanced mathematics or mathematical analysis, linear algebra or advanced algebra. The former focuses on calculus, while the latter focuses on matrix theory and corresponding theories such as linear vector spaces, both of which are the most used fundamentals in the theory of relativity. There are also two basic courses, one is complex variable functions and mathematical physical equations, and the other is probability and statistics; The former is more inclined towards methods and analysis, while the latter is more about understanding and analysis.

In principle, these four mathematics courses are compulsory undergraduate courses in the Department of Physics, and they are very useful for all aspects of physics courses in the Department of Physics. With these foundations, together with the foundation of physics (all courses of general physics plus the four major mechanics), theoretically speaking, it is possible to understand some of the space-time views of special relativity to a certain extent, especially in the theory of electromagnetism. But at this time, you may feel that your understanding of special relativity is only limited to some simple specific situations, and many times there will still be a lot of incomprehension, and after learning these, you still have zero understanding of general relativity.

The following two basic courses are useful for a fundamental understanding of what the theory of relativity is all about, one is modern differential geometry and the other is group theory, but generally these are graduate courses. With the foundation of differential geometry, you can understand what the theory of relativity is talking about (including general relativity) very essentially, and on this basis, if you have the foundation of the group, you can even jump out of the category of relativity, really understand how the theory of gravity is structured, and even understand what the theory of modern physics is based on

In general, that's all the mathematics you need, Advanced Mathematics, Mathematical Analysis, Linear Algebra, Advanced Algebra, Complex Variable Functions and Mathematical Physical Equations, Probability Theory, Differential Geometry, Group Theory. However, the content of mathematics is very wide, and the content listed here is the most closely related to physics, and the knowledge they constitute is also called mathematical physics. But in the end, when learning mathematics, it is best to use the idea of physics to guide, because after all, mathematics is not physics, and vice versa, so I hope you do not lose the essence of physics research in mathematics, which requires you to accumulate enough knowledge of physics accordingly, only in this way can you have a deeper understanding and experience of physics (not mathematics).

Learning Matrix. Actually, the (narrow) theory of relativity doesn't use much mathematics.

The matrix is also only computationally convenient, and it is okay not to use it.

General relativity uses tensors.

It's necessary to learn about tensors.

Special relativity hardly uses new mathematics (meaning newer than calculus).

In the broad sense, differential geometry (which is a higher level of mathematics) is required

If you just want to relativity, you don't need to learn tensor analysis.

The original length of the edge of the cube io, the length of the edge shrinks in the direction of motion as, i = io [1 - u c) 2] (1 2).

The total length of the 4 edges in the direction of motion is, 4 io [1 - u c) 2] (1 2).

The length of the edge in the other directions is unchanged.

The question should be "What is the linear density of B?" ”

First of all, it is easy to obtain =5 3, and in B's view the rod mass increases to m and the length shrinks to l

Then the linear density = m (l ) m l = 25m 9l

It has nothing to do with the m of the rod, and the velocity of A relative to B does not change regardless of whether there is a rod or not. It's like whether you eat with chopsticks or a spoon on the train doesn't matter how fast the train is.

The velocity of B relative to A is the same as the velocity of A relative to B.

Is that the title there? I don't think the question is complete.

ek=mc2-m0c2 m is the moving mass and m0 is the rest mass m0c2=

The relationship between m and m0 is.

m=m0/γ

Under the root number (1-v2 c2).

The result mc2 is about.

Deducting the rest mass, the answer is c

The question is another fool, "event A occurs in the S system x1=50m, t1=2*10 -7 seconds", how did this data come from? Was it observed by a person at the origin of the S system? This is a more certain result, and people who move on S, or those who are not at the origin, are not necessarily this result.

Also, is the time calculated on its own table? This is Einstein's assumption, and Einstein has a derivation loophole, and we can't apply his formula at all. However, during the study period, you can apply the formula to calculate, which shows that you can learn mathematics and understand the theory of relativity.

Think about it, the s' system is in velocity u = along x-x with respect to the s system'Axial forward motion, t=t'=0, x=x'=0, event a occurs in the s system where the person at the origin hears x1=50m, t1=2*10 -7 seconds, event b occurs in the s system where the person at the origin is heard x2=10m, t2=3*10 -7 seconds, then s'What is the time interval between the two events that the surveyor in the department hears? (c is the speed of sound).

Even this question is not the only answer, which shows how many loopholes there are in the questioner.

From the ground, it seems that the length of the isosceles right triangle will be shortened along the direction of velocity, that is, the ruler shrinkage effect, which will change from an isosceles right triangle to an equilateral triangle, and the velocity can only be reduced in the direction of the hypotenuse, that is, the hypotenuse. According to the ruler reduction formula, let l be the original length of the hypotenuse, then the length of the hypotenuse appears to be l, the length of the right-angled side perpendicular to the direction of velocity is unchanged, is l 2, and the length along the direction of velocity is shortened to l 2, because the ground appears to be an equilateral triangle, which is obtained by the Pythagorean theorem.

l 2) 2+(l 2 ) 2=(l ) 2, the solution is 2=3, that is, v= (2 3)c, so choose a. It can also be directly excluded, it is impossible to exceed the speed of light, so exclude cd, and directly select a without calculating

(1) Calculate the quality loss first, and then substitute the mass-energy connection equation to calculate.

2) Multiply the mass by the speed of light squared directly with the formula.