timo

"Factoring out" for the process of going from the left-hand side to the right-hand side. "Distributivity" for the property telling you that you can do that. You did not really describe a process - going from the left-hand side of the equation to the right-hand side was just a guess of what you could have meant.

No. ??? No. Sure. But what you asked for is not what you wanted to ask. What you probably wanted to know is something like "only massless particles can travel at c and only massless particles experience zero time between two different events".

Are there differences between the settings for different groups and if-so then why?

Even though the correct answer to the original and some additional questions are already flying around in this thread, there is so much confusion about so basic things that I find it appropriate to post a little summary of equations and solutions. I'll assume that above [math]\gamma = 7[/math] is correct. See above for the definition of [math]\gamma[/math]. There was no confusion, as far as I see. The rest mass of a proton is [math]m = 938[/math] MeV/c² (off my head, the value might be off by ~5 but that will not matter). The total energy of a (free) proton at rest is [math]E_0 = 938[/math] MeV. This value comes from the mass plugged in into the (in-)famous E=mc². The total energy of a (free) moving proton is [math] E_\gamma = \gamma E_0 [/math]. In this case [math] E_\gamma = 7 \cdot 0.938 \cdot 10^9 \ eV \ = 6.57 \cdot 10^9 [/math] eV. That is pretty much the value you (=gre) had, except that you called it "mass of the proton" and called the unit "Volt" rather than electronvolt. The relativistic mass (perhaps that is what you meant by mass?) of this proton would be [math]6.57 \cdot 10^9 [/math] eV/c². Note the factor c² that make the units match. The total energy is the sum of the kinetic energy and the energy at rest. To get the kinetic energy (what the thread originally was about) you have to substract the rest energy from the total energy, i.e. [math]E_{kin} = E_\gamma - E_0 = 5.63 \cdot 10^9 [/math] eV. This is the step that a lot of people in this thread might have had a problem with. To accelerate a proton to such an energy in an electric field since the proton carries one elementary charge (apart from the sign, depending on definition) the potential difference the proton has to go through -ignoring any effects of losing energy along the way- would then be [math]5.63 \cdot 10^9 [/math] V.

That depends on what you mean: - 6.7e9 Volt definitely is not the proton mass. - It is not even a mass. - It is not an energy, either. - The voltage needed to accelerate a proton in an electric field to 99% of c is not the value stated, either. So it's probably not correct. That does neither give the proton mass (that he said he was looking for but which simply is ~1000 MeV/c²), nor something in units of voltage (what his result was), nor the kinetic energy of a proton at 99% of c (what the thread was about). What it gives is the relativistic mass of an object.

You spelled "Reappear" wrong.

Well, if you consider "my computer" as say the whole part of the machine that is inside my room then physically it read some electromagnetic signals from its surrounding. Ignoring noise, analog-digital conversion, checksums and other technicallities, then logically it read a series of zeroes and ones, yes. However -and I am not sure to what extent this rather fundamental point in my last post became clear- it also read some ASCII characters. Bits and bytes and ASCII characters and hex-numbers are (stuff like "you need 8 bits to form a byte" aside) just different ways to encode the same information. Just like decimal and binary and hexadecimal are simply different ways to express a number; none of the systems being more fundamental than the others. So from "it read bits" or "it read bytes" or "it read ASCII" or "it read hex-code" none of the statements of more correct or fundamental than the other. With "highest level" in my previous post I meant "most human-readable" which is not a really outstanding but a very practical property among a set of equivalent formulations (MC is just assembly translated into numbers ).

My computer read html just a few seconds ago - that's if I count the webbrowser and internet connection being part of the computer (and I do - they are very important parts). So why is the CPU the computer? I dunno why you think so but I will use this definition in the following: If "the computer" was the CPU then on one of the more fundamental levels (not really the most fundamental, but the most fundamental that is reasonable to discuss) the only information that goes through the CPU is time-series of voltages - a completely useless statement . These time-series have a 1-to-1 mapping on larger scales up to assembly code. Above assembly code, the mapping is not 1-to-1; the same C++ program might generate different assembly code on different compilers, for instance. So since by that reasoning (I am only a stupid physicist not a smart computer scientist, so I might be wrong) the highest-level set of expressions that maps 1-to-1 on the intended function of the CPU is assembly, I would chose that as an answer. However: I do disagree with equating "a computer reads" with "a CPU takes as input" - or with the two statements obviously being the same. Formally, the question also had to define whether "only reads X" is true if a computer reads Y and Y is equivalent to X, particularly for the CPU part.

Let's assume we already made all those "ice going to a gaseous state is called sublimation, not boiling"-jokes: You supposedly want to look up the term "latent heat".

Is this homework or can you specify what you mean with a "constant DC current" (a DC current or one that is constant under change of voltage)?

No one (except Bignose) said it would. In part b) in my previous reply I already stated what went wrong, namely that the limit of a function is (or at least "can be considered as") a functional that assigns a value to a function (if the limit exists), not the function itself. Any reason why you (and everyone else) ignored that?

a) The 4 in your case is (using the definition of limits of real-valued functions I know) a number. b) With "it approximates it" and "it is almost equal" you probably mean that for values [math]x \approx 2[/math] the function is almost equal to 4. Strictly speaking that's not what the equation sais. The equation claims that the functional (a functional is an operation that assigns a real value to a function, a typical example being integration) [math] \lim_{x\to 2} (F) [/math] is equal to 4 for [math] F=\frac{x^2 - 4}{x-2} [/math].

I don't think that really affects an MPI school that is aimed towards internationality. The course is -naturally- also quite new. I'm not saying there are no differences (there might as well be differences within the UK in which case the answer boils down to "look it up at the institutes you are interested in"). I just would not expect any for this rather broad question. Or to approach your statement from the other side: How and why would you think differences between British and German university system would have a stronger impact on the question at hand than the differences between UK university A and UK university B?

Speaking for the MPI Göttingen (german, but I expect it to be similar to UK): Your best bet is browsing a few pages of institutes and see what their requirements are.

One of the most important aspects of fitting functions is realizing that "I want to fit a function through a number of points" is not a sufficient statement for a unique solution to exist (or make sense). You have to specify what kind of function (ideally described by a small amount of free parameters that then shall be fitted) you want to fit to the data.

So how do these textbooks justify equating the p with mc and what do these letters stand for? It looks completely ridiculous to me.

I'm not sure if it is too obvious for you but perhaps I should nit-pick a bit: If he really said that then your teacher is wrong. Bodies maintaining their states of rest or uniform motion unless there is a force acting on them is supposedly always true. What does change at velocities close to the speed of light is that Newton's prediction of the reaction of a particle on a force becomes bad/wrong.

Two possible answers: 1) Why should it? 2) That's what the distributive property of the real numbers is like: 3x + 2x = (3+2)x = 5x hence 8a^5 + 10a^5 = (8 + 10)a^5 = 18a^5.

Well, then: I am a physicist. I am open minded by your definition. Example: I was receptive to the new idea of quantum mechanics up to the point that I passed the quantum mechanics exam. I would claim this statement applies to pretty much every physicist.

By realizing that it is a quadratic equation in x² that you probably know how to find all solutions for.

42.

I don't think I've ever had a party's or candidate's proficiency or attitude towards foreign policy playing a major role for my voting behavior. How about you? I would expect that national issues overshadow anything else in a presidential campaign (or even outside of campaigns that national issues are -naturally- seen as the most important). Also, while Mrs. Palin did demonstrate quite a lack of competence when measured by the standards you might expect from a political leader there's one thing that might put this lack of knowledge/competence in scale: She possibly still knows at least as much about foreign countries/politics as the average voter.

I would imagine it being more or less defined. For example, you can easily show that if reals > 0 form a group under multiplication, then for the extension including the negative reals to form a group under multiplication the product of two negative numbers (particularly a negative number and its inverse) must be positive (=1 in the special case). It's not clear to me to what extent you can see this example I made up as an axiom (is there a reason to assume either set forms a group under multiplication?) or a proof (if you assume they should form a group under multiplication then the property you wanted follows). You probably can make up quite a lot of examples like the one I just presented. I would be interesting to hear a canonic viewpoint (or whether one exists).

"Most" is absolutely sufficient to say that the spin-zeroness of the photons is not the reason that there is no anti-photon. "There are photons with spin 1" would have been the better wording, perhaps. As it's already been said, the post I was referring to has been edited afterwards, hence the statement is a bit out of context.

And in 3D LA should be sufficient, alone